A convex lens of refractive index 3/2 has a power of `2.5^(@)`. If it is placed in a liqud of refractive index 2,the new power of the lens is

Options

A. 2.5D
B. `-2.5D`
C. 1.25D
D. `-1.25D`

Explanation

To solve the problem step by step, we need to find the new power of a convex lens when it is placed in a liquid of a different refractive index.

### Step 1: Understand the given data
We have:
- Refractive index of the lens (convex lens), \( \mu_g = \frac{3}{2} \)
- Power of the lens in air, \( P_1 = 2.5 \, D \) (diopters)
- Refractive index of the liquid, \( \mu_l = 2 \)

### Step 2: Calculate the effective refractive index of the lens in the liquid
The effective refractive index of the lens when placed in a liquid can be calculated using the formula:
\[ \mu_{new} = \frac{\mu_g}{\mu_l} \]
Substituting the values:
\[ \mu_{new} = \frac{\frac{3}{2}}{2} = \frac{3}{4} \]

### Step 3: Use the formula for power of the lens
The power of the lens in a medium is given by:
\[ P = \frac{\mu_{new} - 1}{\mu_g - 1} \times P_1 \]
Where:
- \( P_1 \) is the initial power of the lens in air.
- \( \mu_{new} \) is the effective refractive index in the liquid.
- \( \mu_g \) is the refractive index of the lens.

### Step 4: Substitute the values into the power formula
Substituting the values we have:
\[ P_2 = \frac{\frac{3}{4} - 1}{\frac{3}{2} - 1} \times 2.5 \]

### Step 5: Simplify the expression
Calculate \( \frac{3}{4} - 1 \) and \( \frac{3}{2} - 1 \):
\[ \frac{3}{4} - 1 = \frac{3}{4} - \frac{4}{4} = -\frac{1}{4} \]
\[ \frac{3}{2} - 1 = \frac{3}{2} - \frac{2}{2} = \frac{1}{2} \]
Now substitute these back into the power formula:
\[ P_2 = \frac{-\frac{1}{4}}{\frac{1}{2}} \times 2.5 \]

### Step 6: Calculate the new power
Calculating the fraction:
\[ \frac{-\frac{1}{4}}{\frac{1}{2}} = -\frac{1}{4} \times \frac{2}{1} = -\frac{1}{2} \]
Now multiply by the power:
\[ P_2 = -\frac{1}{2} \times 2.5 = -1.25 \, D \]

### Final Answer
The new power of the lens when placed in the liquid is:
\[ \boxed{-1.25 \, D} \]